Question

An ideal ramjet is to operate with exhaust gases optimally expanded to ambient pressure at an altitude where temperature is 220 K. The exhaust speed at the nozzle exit is 1200 m/s at a temperature of 1100 K.
Given: \( \gamma = 1.4 \) at 220 K; \( R = 287 \, \text{J/(kg-K)} \) for air; \( \gamma = 1.33 \) at 1100 K; \( R = 287 \, \text{J/(kg-K)} \) for exhaust gases.
The cruise speed of this ramjet is _________ m/s (rounded off to nearest integer).

Hint:

The cruise speed of a ramjet can be estimated using the temperature and gas constant at the exhaust, along with the thermodynamic properties of the exhaust gases and air.

Answer 1

Correct Answer: 545

For an ideal ramjet, the relation between the exhaust velocity and the cruise speed is derived from the conservation of energy and momentum for ideal gas flow. The equation for the exit velocity of the exhaust gases is: \[ V_e = \sqrt{\frac{2 \gamma R T_e}{\gamma - 1} \left( 1 - \left( \frac{P_e}{P_0} \right)^{\frac{\gamma - 1}{\gamma}} \right)}, \] where \( T_e = 1100 \, \text{K} \) is the exhaust temperature, \( R = 287 \, \text{J/(kg-K)} \) is the specific gas constant, and \( \gamma = 1.33 \). Given the exhaust velocity \( V_e = 1200 \, \text{m/s} \), we need to calculate the ramjet cruise speed.
The exhaust velocity at cruise is related to the difference in pressure and velocity: \[ V_{\text{cruise}} = \sqrt{2 \times R \times T_0 \times \gamma}, \] where \( T_0 = 220 \, \text{K} \) is the initial temperature, and \( \gamma = 1.4 \). After solving for the cruise speed, we find: \[ \boxed{550 \, \text{m/s}}. \]